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I was thinking about the problem which says:

Let $f(z)$ be a function defined by: $$f(z)=\frac{\sin(1/z)}{z^{2}+11z+13}.$$ Then which of the following is correct?

(a) No singularity.

(b) Only poles.

(c) Only an essential singularity.

(d) Both an essential singularity and poles.

I have eliminated the possibilities of (a) and (b). But can not decide on (c) and (d). Clearly, $f(z)$ has essential singularity as can be explained by expanding $\sin(1/z)$ whose principal part contains infinite number of terms. But can not decide about option (d) which says about the existence of both poles and essential singularity. It will be helpful if someone throws light on it..Thanks in advance.

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When is the denominator $0$? –  Nameless Nov 30 '12 at 18:14
    
@Nameless Yes,i have also thought about it. We can simply solve z^2++11z+13=0.But what will happen if we expand 1/(z^2+11z+13) in an infinite series form? –  learner Nov 30 '12 at 18:17
    
At what point ? –  Nameless Nov 30 '12 at 18:23
    
$1/(z^2+11z+13)$ $=\frac{1}{13} - \frac{11}{13^2} z + \frac{-13+11^2}{13^3} + \cdots$. The radius of convergence is $\left|\left(-11+\sqrt{69}\,\right)/2\right|$, since that's the distance from $0$ to the nearest pole of that rational function. I'm not sure what you had in mind doing with the power series. –  Michael Hardy Nov 30 '12 at 19:45
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1 Answer 1

up vote 2 down vote accepted

There are poles at the two roots of the quadratic polynomial in the denominator. There is an essential singularity at $0$, since $\lim\limits_{z\to0} f(z)$ doesn't exist within $\mathbb C \cup \{\infty\}$.

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